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Thread: Fourier series and (possibly) Parsevals Theorem

  1. #1
    Super Member Deadstar's Avatar
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    Fourier series and (possibly) Parsevals Theorem

    Let $\displaystyle f$ be the function given by $\displaystyle f(\theta) = |\theta|$ for $\displaystyle \theta\in [ -\pi , \pi ]$.

    a) Calculate the fourier coefficients of $\displaystyle f$ and show that...

    $\displaystyle \hat{f}(n) = \frac{(-1)^n - 1}{\pi n^2}$ for $\displaystyle n \neq 0$

    $\displaystyle \hat{f}(n) = \frac{\pi}{2}$ for $\displaystyle n = 0$.

    done...

    b) What is the Fourier series of $\displaystyle f$ in terms of sines and cosines?

    My answer, $\displaystyle \frac{\pi}{2} + \sum_{n=-\infty}^{\infty} \frac{(-1)^n - 1}{\pi n^2} \cos(n\theta)$.

    c) Taking $\displaystyle \theta= 0$, prove that

    $\displaystyle \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{\pi^2}{8}$ and $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

    Now this, I suspect, should have just been a simple application of Parsevals theorem but I'm going wrong somewhere.

    Parsevals Theorem.

    $\displaystyle \left ( \frac{1}{2 \pi} \int_{-\pi}^{\pi} |f(\theta)|^2 d \theta \right)^{\frac{1}{2}} = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}}$

    But with $\displaystyle \theta = 0$ the left hand side is 0.

    Thus we get,

    $\displaystyle 0 = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}}$

    = $\displaystyle \left ( \frac{\pi^2}{4} + \sum_{n=-\infty}^{\infty} \left (\frac{1-1}{n^2 \pi} \right )^2 \right)^{\frac{1}{2}} = \frac{\pi}{2}$

    So clearly I'm doing something wrong... Most likely in that last 2 steps...
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    Let $\displaystyle f$ be the function given by $\displaystyle f(\theta) = |\theta|$ for $\displaystyle \theta\in [ -\pi , \pi ]$.

    a) Calculate the fourier coefficients of $\displaystyle f$ and show that...

    $\displaystyle \hat{f}(n) = \frac{(-1)^n - 1}{\pi n^2}$ for $\displaystyle n \neq 0$

    $\displaystyle \hat{f}(n) = \frac{\pi}{2}$ for $\displaystyle n = 0$.

    done...

    b) What is the Fourier series of $\displaystyle f$ in terms of sines and cosines?

    My answer, $\displaystyle \frac{\pi}{2} + \sum_{n=-\infty}^{\infty} \frac{(-1)^n - 1}{\pi n^2} \cos(n\theta)$.


    For even n's the summands of this series are zero, so you can take only odd n's in the sum and since you have a square in the denominator this

    is the same as twice the sum over the odd natural numbers (taking $\displaystyle \theta=0$):

    $\displaystyle \frac{\pi}{2} + \sum_{n=-\infty}^{\infty} \frac{(-1)^n - 1}{\pi n^2} \cos(n\cdot 0)=\frac{\pi}{2}+\frac{2}{\pi}\sum^\infty_{k=0}\fr ac{-2}{(2k-1)^2}$ $\displaystyle = \frac{\pi}{2}-\frac{4}{\pi}\sum^\infty_{k=0}\frac{1}{(2k-1)^2}$ .

    Now equal the above to zero and you'll get what you want...

    Tonio



    c) Taking $\displaystyle \theta= 0$, prove that

    $\displaystyle \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{\pi^2}{8}$ and $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

    Now this, I suspect, should have just been a simple application of Parsevals theorem but I'm going wrong somewhere.

    Parsevals Theorem.

    $\displaystyle \left ( \frac{1}{2 \pi} \int_{-\pi}^{\pi} |f(\theta)|^2 d \theta \right)^{\frac{1}{2}} = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}}$

    But with $\displaystyle \theta = 0$ the left hand side is 0.

    Thus we get,

    $\displaystyle 0 = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}}$

    = $\displaystyle \left ( \frac{\pi^2}{4} + \sum_{n=-\infty}^{\infty} \left (\frac{1-1}{n^2 \pi} \right )^2 \right)^{\frac{1}{2}} = \frac{\pi}{2}$

    So clearly I'm doing something wrong... Most likely in that last 2 steps...
    .
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